Solve: sqrt(x^2 - 3x - 6) - 3 = x - 2

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Solve: sqrt(x^2 - 3x - 6) - 3 = x - 2

Mathematics
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There is no solution to this. If it were \[\pm \sqrt{x ^{2} - 3x - 6} = x + 1\] Then the solution to the negative part would be (-7/5)
Why do you say, that there's no solution? U just added 3 to both sides, so your equation isn't really any different from MargueriteSperduto's.
wait so whats the answer?

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His answer is right, it's just wrong, that he says that there's no solution. \[\sqrt{x^2-3x-6} -3 = x-2\] If you're adding 3 to both sides you get this: \[\sqrt{x^2 -3x-6} = x+1\] Now you're squaring both sides: \[x^2-3x-6=(x+1)^2 \] This also equals: \[x^2-3x-6=x^2+2x+1\] Now you subscrat x^2 on both sides: \[-3x -6 = 2x+1\] Adding 3x on both sides gets you this: \[-6 = 5x+1\] Now you're subscrating one on both sides:\[-7=5x\] Et voila: \[x= \frac{-7}{5}\]
From an algebraic sense, you are correct there is a solution. However, in your steps to finding that answer, you had to square both sides which changes elements of the domain because of the square root. If you had checked your answer, you would indeed see that: \[\sqrt{x^2 -3x - 6} = x+1\] when \[x = (-7/5)\], gives you, \[(-2/5) = (2/5)\], hence why there is no real solution and the +- is necessary
And if you still don't believe me, let our genius friend explain: http://bit.ly/gJaIjy

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